Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-3x+9y &= -3 \\ -2x-6y &= 8\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-6y = 2x+8$ Divide both sides by $-6$ to isolate $y$ $y = {-\dfrac{1}{3}x - \dfrac{4}{3}}$ Substitute this expression for $y$ in the first equation. $-3x+9({-\dfrac{1}{3}x - \dfrac{4}{3}}) = -3$ $-3x - 3x - 12 = -3$ Simplify by combining terms, then solve for $x$ $-6x - 12 = -3$ $-6x = 9$ $x = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $x$ back into the top equation. $-3( -\dfrac{3}{2})+9y = -3$ $\dfrac{9}{2}+9y = -3$ $9y = -\dfrac{15}{2}$ $y = -\dfrac{5}{6}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = -\dfrac{5}{6}$.